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Chapter 42: Problem 36

Electrons with a maximum KE of \(3.00 \mathrm{eV}\) are ejected from a metal surface by ultraviolet radiation of wavelength \(150 \mathrm{~nm}\). Determine the work function of the metal, the threshold wavelength of the metal, and the retarding potential difference required to stop the emission of electrons.

Short Answer

Expert verified
Work function: 5.28 eV, Threshold wavelength: 235 nm, Retarding potential: 3.00 V.

Step by step solution

01

Understanding the Photoelectric Effect Equation

The photoelectric effect equation is given by \( E = W + KE_{max} \), where \( E \) is the energy of the incident photons, \( W \) is the work function of the metal, and \( KE_{max} \) is the maximum kinetic energy of the ejected electrons. The energy of the incident photons is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \), \( c \) is the speed of light \( (3.00 \times 10^8 \text{ m/s}) \), and \( \lambda \) is the wavelength of the light.
02

Calculate Incident Photon Energy

First, calculate the energy of the photons using the given wavelength \( \lambda = 150 \text{ nm} = 150 \times 10^{-9} \text{ m} \). Substituting \( h \), \( c \), and \( \lambda \) into the equation: \[ E = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{150 \times 10^{-9} \text{ m}} = 1.326 \times 10^{-18} \text{ J} \]Convert this to electron volts for convenience: \[ E = \frac{1.326 \times 10^{-18} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 8.28 \text{ eV} \].
03

Determine the Work Function

Now, use the equation \( E = W + KE_{max} \) to solve for the work function \( W \):\[ W = E - KE_{max} = 8.28 \text{ eV} - 3.00 \text{ eV} = 5.28 \text{ eV} \].Thus, the work function of the metal is \( 5.28 \text{ eV} \).
04

Calculate the Threshold Wavelength

The threshold wavelength \( \lambda_0 \) is related to the work function \( W \) and is given by \( W = \frac{hc}{\lambda_0} \). Rearrange to solve for \( \lambda_0 \):\[ \lambda_0 = \frac{hc}{W} = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{5.28 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}} = 235 \text{ nm} \].
05

Calculate the Retarding Potential Difference

The retarding potential difference \( V \) required to stop the emission of electrons can be found using the relationship \( V = \frac{KE_{max}}{q} \), where \( q \) is the electron charge \( (1.602 \times 10^{-19} \text{ C}) \):\[ V = \frac{3.00 \text{ eV}}{1.00 \text{ e}} = 3.00 \text{ V} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Function
The work function is a crucial concept in understanding the photoelectric effect, which explains how light can cause electrons to be ejected from a material. Think of the work function as the minimum amount of energy needed to remove an electron from the surface of a metal. This is important because, for an electron to be emitted, the energy of the incoming light must exceed this threshold.

In mathematical terms, the work function (\( W \)) is given in electron volts (eV). For the problem we're discussing, the energy required is calculated using the formula \( E = W + KE_{max} \), where \( KE_{max} \) is the maximum kinetic energy of the electrons that are ejected.

  • Here, the incoming photon energy is \( 8.28 \) eV, and the kinetic energy of the emitted electrons is \( 3.00 \) eV.
  • This tells us that the work function for this metal is \( W = E - KE_{max} = 5.28 \) eV.
This value essentially indicates how tightly the electrons are bound to the atom in the metal, and it varies from one material to another.
Threshold Wavelength
The threshold wavelength is the maximum wavelength of light that can still cause electron ejection in the photoelectric effect. It is directly linked to the work function of the material. Essentially, it represents the light's energy just sufficient enough to remove an electron from the metal surface.

The formula linking the work function and the threshold wavelength (\( \lambda_0 \)) is \( \lambda_0 = \frac{hc}{W} \):

  • \( h \) is Planck's constant (\(6.626 \times 10^{-34} \) Js)
  • \( c \) is the speed of light (\(3.00 \times 10^8 \) m/s)
For this exercise, we can find the threshold wavelength by substituting the work function into the equation:

  • Using \( W = 5.28 \) eV, converting energy to joules (by multiplying with \( 1.602 \times 10^{-19} \) J/eV), gives us \( \lambda_0 = 235 \) nm.
This means electromagnetic radiation with a wavelength longer than 235 nm will not have enough energy to dislodge electrons from the surface of this metal.
Retarding Potential Difference
In the context of the photoelectric effect, the retarding potential difference is the voltage required to stop the electrons from moving. If the emitted electrons have a certain amount of kinetic energy, applying a reverse electric potential can stop them, effectively controlling the current of ejected electrons.

The retarding potential difference (\( V \)) is calculated using the formula \( V = \frac{KE_{max}}{q} \):

  • \( KE_{max} \) refers to the maximum kinetic energy of the electrons. In our case, this is \( 3.00 \) eV.
  • \( q \) is the electron charge (\( 1.602 \times 10^{-19} \) C).
By substituting the kinetic energy into the formula, we find:

  • The retarding potential required to stop the electrons is \( V = 3.00 \) V.
This value helps in measuring the kinetic energy of the photoelectrons and is crucial for experiments where precise control over electron flow is necessary.

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Most popular questions from this chapter

In a process called pair production, a photon is transformed into an electron and a positron. A positron has the same mass \(\left(m_{e}\right)\) as the electron, but its charge is \(+e\). To three significant figures, what is the minimum energy a photon can have if this process is to occur? What is the corresponding wavelength? The electron-positron pair will come into existence moving with some minimum amount of KE. The particles will separate, and as they do they will slow down. When far apart each will have a mass of \(9.11 \times 10^{-31} \mathrm{~kg} .\) In effect, KE goes into \(\mathrm{PE}\), which is manifested as mass. Thus, the minimum energy photon at the start of the process must have the energy equivalent of the free-particle mass of the pair at the end of the process. Hence,

Suppose that a 3.64-nm photon moving in the \(+x\) -direction collides head-on with a \(2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) electron moving in the \(-x\) direction. If the collision is perfectly elastic, find the conditions after collision. From the law of conservation of momentum, Momentum before = Momentum after $$ \frac{h}{\lambda_{0}}-m v_{0}=\frac{h}{\lambda}-m v $$ But, from \(\underline{\text { Problem } 42.9,} h / \lambda_{0}=m u\) in this case. Hence, \(h / \lambda=m v\). Also, for a perfectly elastic collision, $$ \begin{array}{l} \text { KE before }=\text { KE after } \\ \frac{h c}{\lambda_{0}}+\frac{1}{2} m v_{0}^{2}=\frac{h c}{\lambda}+\frac{1}{2} m v^{2} \end{array} $$ Using the facts that \(h / \lambda_{0}=m v_{0}\) and \(h / \lambda=m v\), we find $$ v_{0}\left(\mathrm{c}+\frac{1}{2} v_{0}\right)=v\left(\mathrm{c}+\frac{1}{2} v\right) $$ Therefore, \(v=\mathrm{u}_{0}\) and the electron moves in the \(+\chi\) -direction with its original speed. Because \(h / \lambda=m v=m u_{0}\), the photon also "rebounds," and with its original wavelength.

To break a chemical bond in the molecules of human skin and thus cause sunburn, a photon energy of about \(3.50 \mathrm{eV}\) is required. To what wavelength does this correspond? $$ \lambda=\frac{h \mathrm{c}}{\mathrm{E}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{(3.50 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}=354 \mathrm{~nm} $$ Ultraviolet radiation causes sunburn.

Determine the energy in joules of a photon that has a wavelength of \(589.3 \mathrm{~nm}\) at the center of the sodium doublet.

II] A particle of mass \(m\) is confined to a circular orbit with radius \(R\). For resonance of its de Broglie wave on this orbit, what energies can the particle have? Determine the KE for an electron with \(R=\) \(0.50 \mathrm{~nm}\). To resonate on a circular orbit, a wave must circle back on itself in such a way that crest falls upon crest and trough falls upon trough. One resonance possibility (for an orbit circumference that is four wavelengths long) is shown in \(\underline{\text { Fig. } 42-2 . \text { In general, resonance }}\) occurs when the circumference is \(n\) wavelengths long, where \(n=\) \(1,2,3, \ldots .\) For such a de Broglie wave $$ n \lambda_{n}=2 \pi R \quad \text { and } \quad p_{n}=\frac{h}{\lambda_{n}}=\frac{n h}{2 \pi R} $$ Fig. \(42-2\) As in Problem \(42.17\), $$ (\mathrm{KE})_{n}=\frac{p_{n}^{2}}{2 m}=\frac{n^{2} h^{2}}{8 \pi^{2} R^{2} m} $$ The energies are obviously quantized. Placing in the values requested leads to $$ (\mathrm{KE})_{n}=2.4 \times 10^{-20} n^{2} \mathrm{~J}=0.15 n^{2} \mathrm{eV} $$
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